In A, B, C are having some marbles with each of them. A has given B and C the same number of marbles each of them already have. Then, B gave C and A the same number of marbles they already have. Then C gave A and B the same number of marbles they already have. At the end A, B, and C have equal number of marbles. If x, y, z are the marbles initially with A, B, C respectively. If the total number of marbles are 72, then the number of marbles with A at the startingA. 20B. 30C. 32D. 39

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Jun 21st, 2005
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Moulana Shareef

Dec 23rd, 2005

total no of marbles=72
at the end all are having equal no of marbles i.e.,each having 24.
calculate from bottom:
A        B         C
39      21        12 (A gave 21 to B and 12 to C)
6        42        24 (B gave 6 to A and 12 to C)
12       12        48 ( c gave 12 to A and 12 to B)
24      24         24
ans:39

swati shukla

Mar 14th, 2007

Can be calculated by linear eqns. Where you get x+2y=87. Then x has a value such that y should be integral. Substitute the options. 39 is the right answer

vikasamiksha
Profile Answers by vikasamiksha

Jan 13th, 2010

At the end they have equal no of marbels that is 72/3=24
C gave same no. of marbels to A & B at the end they have 24 so C gave 12 to each of them so A,B,C will be having 12,12,48
B gave same no. of marbels to A & C they have , so A,B,C will be having 6,12+24+6,24 ie 6,42,24
A gave same no. of marbels they have, so they will be having 6+21+12,21,12 ie 39,21,12
The only thing is that you have to start it from the end the condition that is given that they have same no of marbles at the end.